Annals of Mathematics, 150 (1999), 283–311
The number of solutions of φ (x) = m
By Kevin Ford
Abstract
An old conjecture of Sierpiński asserts that for every integer k > 2, there
is a number m for which the equation φ(x) = m has exactly k solutions. Here
φ is Euler’s totient function. In 1961, Schinzel deduced this conjecture from his
Hypothesis H. The purpose of this paper is to present an unconditional proof
of Sierpiński’s conjecture. The proof uses many results from sieve theory, in
particular the famous theorem of Chen.
1. Introduction
Of fundamental importance in the theory of numbers is Euler’s totient
function φ(n). Two famous unsolved problems concern the possible values
of the function A(m), the number of solutions of φ(x) = m, also called the
multiplicity of m. Carmichael’s Conjecture ([1],[2]) states that for every m,
the equation φ(x) = m has either no solutions x or at least two solutions. In
other words, no totient can have multiplicity 1. Although this remains an open
problem, very large lower bounds for a possible counterexample are relatively
10
easy to obtain, the latest being 1010 ([10, Th. 6]). Recently, the author has
also shown that either Carmichael’s conjecture is true or a positive proportion
of all values m of Euler’s function give A(m) = 1 ([10, Th. 2]). In the 1950’s,
Sierpiński conjectured that all multiplicities k > 2 are possible (see [9] and
[16]), and in 1961 Schinzel [17] deduced this conjecture from his well-known
Hypothesis H [18].
Schinzel’s Hypothesis H. Suppose f1 (n), . . . , fk (n) are irreducible,
integer-valued polynomials (for integral n) with positive leading coefficients.
Also suppose that for every prime q, q - f1 (n) · · · fk (n) for some n. Then the
numbers f1 (n), . . . , fk (n) are simultaneously prime for infinitely many positive
integers n.
Hypothesis H is a generalization of Dickson’s prime k-tuples conjecture [7],
which covers the special case when each fi is linear. Schinzel’s construction of
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KEVIN FORD
a number with multiplicity k requires that as many as k distinct polynomials
with degrees as large as 3k simultaneously represent primes. Using a different,
iterative argument, the author ([10, Th. 9]) deduced Sierpiński’s Conjecture
from the Prime k-tuples Conjecture, the argument requiring only three linear
polynomials to be simultaneously prime. The proof is quite simple, and we
reproduce it here.
Lemma 1.1. Suppose A(m) = k and p is a prime satisfying
(i) p > 2m + 1,
(ii) 2p + 1 and 2mp + 1 are prime,
(iii) dp + 1 is composite for all d | 2m except d = 2 and d = 2m.
Then A(2mp) = k + 2.
Proof. Suppose φ(x1 ) = · · · = φ(xk ) = m and φ(x) = 2mp. Condition (i)
implies p - x, hence p | (q − 1) for some prime q dividing x. Since (q − 1) | 2mp,
we have q = dp + 1 for some d | 2m. We have q > 2p, so q 2 - x and thus
φ(x) = (q − 1)φ(x/q). By conditions (ii) and (iii), either q = 2p + 1 or
q = 2mp+1. In the former case, φ(x/q) = m, which has solutions x = (2p+1)xi
(1 6 i 6 k). In the latter case, φ(x/q) = 1, which has solutions x = q and
x = 2q.
Now suppose A(m) = k and let d1 , . . . , dj be the divisors of 2m with
3 6 di < 2m. Let p1 , . . . , pj be distinct primes satisfying pi > di for each i.
Using the Chinese Remainder Theorem, let a mod b denote the intersection
mod pi (1 6 i 6 j). Then for every h and i,
of the residue classes −d−1
i
(a + bh)di + 1 is divisible by pi , hence composite for large enough h. The
Prime k-tuples Conjecture implies that there are infinitely many numbers h so
that p = a + hb, and 2p + 1 and 2mp + 1 are simultaneously prime. By Lemma
1.1, A(2mp) = k + 2 for each such p. Starting with A(1) = 2 and A(2) = 3,
Sierpiński’s Conjecture follows by induction on k.
Although Hypothesis H has not been proved in even the simplest case of
two linear polynomials (generalized twin primes), sieve methods have shown
the conclusion to hold if the numbers f1 (n), . . . , fk (n) are allowed to be primes
or “almost primes” (nonprimes with few prime factors). See [13] for specific
statements. Recently, the author and S. Konyagin [11] employed the theory of
almost primes to prove that A(m) = k is solvable for every even k. Although
falling short of a complete proof of Sierpiński’s Conjecture, the method did
provide a solution to the corresponding problem for the sum of divisors function
σ(n). It is proved ([11], Theorem 1) that for every k > 0, there is a number
m for which σ(x) = m has exactly k solutions, settling another conjecture
of Sierpiński. Curiously, the barrier to solving A(m) = k for odd k by this
method is the probable nonexistence of a number with A(m) = 1.
THE NUMBER OF SOLUTIONS OF φ(X)=M
285
Combining a different application of almost primes with the iterative approach from [10], we now provide an unconditional proof of Sierpiński’s Conjecture.
Theorem 1. For each integer k
> 2, there is an integer m with A(m) = k.
Our proof of Theorem 1 is by induction on k. We start with A(10) = 2,
A(2) = 3 and for the induction step, we show:
Theorem 2. Suppose A(m) = k and m is even. Then there is a number
m0 such that A(mm0 ) = k + 2.
Combining Theorem 1 with Theorem 2 of [10] gives a strong result on the
distribution of totients (values taken by φ(n)) with a given multiplicity.
Corollary 3. For each k > 2, a positive proportion of all totients have
multiplicity k. Specifically, if V (x) is the number of totients 6 x, and Vk (x) is
the number of totients 6 x with multiplicity k, then for each k > 2,
lim inf
x→∞
Vk (x)
> 0.
V (x)
At the center of the proof of Theorem 2 is a famous theorem of Chen ([3],
[4]).
Lemma 1.2 (Chen’s theorem). For each even natural number m and
x > x0 (m), there are À x/ log2 x primes s ∈ (x/2, x], s ≡ 1 (mod m) such
that (s − 1)/m has at most two prime factors, each of which exceeds x1/10 .
Remark. A proof of Chen’s Theorem also appears in Chapter 11 of [13].
What Chen actually proved is that there are À x/ log2 x primes p 6 x with
p + 2 having at most two prime factors. Trivial modifications to the proof give
Lemma 1.2.
Hypothesis S(m). For some constants c > 0 and x0 , if x > x0 then the
number of primes s ∈ (x/2, x] for which (s−1)/m is also prime is > cx/ log2 x.
We suppose φ(α) = m has the k solutions α1 , . . . , αk . The proof is then
broken into two parts: (I) assuming Hypothesis S(m) is true; (II) assuming
Hypothesis S(m) is false. For Part I we show that for many such primes s,
A(mq(q − 1)) = k + 2, where q = (s − 1)/m and φ(n) = mq(q − 1) has the
k + 2 solutions n = q 2 αi (i = 1, . . . , k), n = q(mq + 1) and n = 2q(mq + 1)
(the “trivial” solutions).
For Part II, Chen’s Theorem tells us that for an unbounded set of x,
there are À x/ log2 x primes s ∈ (x/2, x] with s = 1 + mpq, where p and
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KEVIN FORD
q are primes and x1/10 6 p 6 q 6 x9/10 . We show that for many such s,
A(mpq(p − 1)(q − 1)) = k + 2, the solutions of
φ(n) = mpq(p − 1)(q − 1)
(1.1)
being only the “trivial” ones n = p2 q 2 αi (i = 1, . . . , k), n = pq(mpq + 1) and
n = 2pq(mpq + 1). Proving this is more difficult than Part I, but a simple
heuristic indicates that such s are likely. The basis for the heuristic (and the
actual proof) is an extension of the classical result of Hardy and Ramanujan
[14] that most integers n 6 x have about log log x prime factors. Erdős [8]
combined their method with Brun’s sieve to show that for most primes p 6 x,
p − 1 has about log log x prime factors. One can further refine these ideas (e.g.
Lemma 2.8 of [10] or Lemma 4.3 below) to show that for most primes p 6 x,
the prime factors of p − 1 are nicely distributed. This means that p − 1 has
about log log b − log log a prime factors in the interval [a, b] for 1 6 a 6 b 6 x.
Call such primes “normal”. If p and q are both normal primes and x is large,
then w = p(p − 1)q(q − 1)m has about 2 log log x prime factors. Fix t and
suppose n has t distinct prime factors r1 , . . . rt , different from p and q. For
simplicity suppose n is square-free. Then
φ(n) = (r1 − 1) · · · (rt − 1) = w.
For a given factorization w = d1 · · · dt , the probability that d1 + 1, . . . , dt + 1
are all prime is about ((log d1 ) · · · (log dt ))−1 . The critical case is t = 2. Since
p and q are normal, w has about 2 log log y prime factors 6 y, and thus has
about (log y)2 log 2 divisors 6 y. By partial summation,
X
w=d1 d2
di >2
1
¿ (log x)2 log 2−2 ,
(log d1 )(log d2 )
the right side representing the probability that (1.1) has a nontrivial solution
for a given pair (p, q). Similarly, it can be shown that
X
w=d1 ···dt
di >2
1
¿ (log x)2 log 2−2 ,
(log d1 ) · · · (log dt )
the bulk of the sum coming from taking d1 , . . . , dt−2 all small.
The biggest practical problem with these arguments is the fact that φ(n)
is quadratic in q for Part I and quadratic in both p and q for Part II. Thus,
when certain sieve estimates are employed, the bounds have factors of the form
µ ¶¶
Yµ
1 D
,
1−
P (D, z) =
p p
p6z
THE NUMBER OF SOLUTIONS OF φ(X)=M
287
where ( D
p ) is the Legendre symbol. Heuristically, such products should be
O(1) on average, and showing this for a wide range of the parameters D and z
requires strong results on the distribution of primes in arithmetic progressions
(see (2.1) and Lemmas 2.6 – 2.8 and 4.2 below).
In the next section, various lemmas needed in the proof of both parts are
presented. Section 3 is devoted to the proof in Part I. Section 4 contains further
lemmas needed for Part II, and the argument counting nontrivial solutions of
(1.1) with (n, pq) > 1. The last section deals with the hardest case, namely
counting solutions of (1.1) with (n, pq) = 1.
The following notation will be used throughout. Let P + (n) denote the
largest prime factor of n and P − (n) denote the smallest prime factor of n.
The functions ω(n) and Ω(n) count the number of distinct prime factors of
n, respectively the number of prime factors counted according to multiplicity.
Let Ω(n, U, T ) denote the number of prime
¡ a ¢ factors p of n such that U 6 p < T ,
counted according to multiplicity. Let n denote the Legendre-Jacobi symbol
and let τ (n) be the number of positive divisors of n. Constants implied by the
Landau O−, Vinogradov ¿ − and À − symbols may depend on m, but not
on any other parameter unless indicated. The symbols p, q, and $ will always
denote primes. The symbol m will always refer to the number in Theorem 2.
The cardinality of a set S will be denoted as #S or |S|.
2. Preliminary lemmas
Aside from Chen’s Theorem, the other sieve results we require all follow
from a basic sieve inequality called the “Fundamental Lemma” of the combinatorial sieve (Theorem 2.5 of [13]).
Lemma 2.1. Suppose F is a polynomial of degree h > 1 with integer
coefficients and positive leading coefficient. Denote by %($) the number of
solutions of F (n) ≡ 0 (mod $). For 2 6 z 6 x,
¶
Yµ
%($)
−
1−
#{n 6 x : P (F (n)) > z} = x
$
$ 6z
n
³ √
´o
× 1 + Oh e− log x + e−u(log u−log log 3u−log h−2) ,
where u = log x/ log z.
Proof. If %($) = $ for some prime $ 6 z, then the left side is zero.
Otherwise the lemma follows from [13, Th. 2.6] and following remarks.
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KEVIN FORD
Remark. The error term is Oh (1), and this will suffice for most of our
applications. Also, for future reference, we note that for $ - 2a,
µ 2
¶
b − 4ac
(2.1)
#{n : an2 + bn + c ≡ 0 (mod $)} = 1 +
.
$
Lemma 2.2. Suppose
F (n) = (a1 n + b1 ) · · · (ah n + bh ),
where the ai are positive integers and the bi are integers. Suppose
Y
E := a1 · · · ah
(ai bj − aj bi ) 6= 0.
16i<j 6h
If 2 6 z
6 x and log E 6 log3 x, then
x
#{n 6 x : P (F (n)) > z} ¿h
(log z)h
−
µ
E
φ(E)
¶h
¿h x(log log x)h (log z)−h .
Proof. In the notation of Lemma 2.1, we have %($) = h when $
therefore the number of n in question is
¿h x
Yµ
$ 6z
$ -E
1
1−
$
µ
¶h
¿h x
E
φ(E)
¶h
-
E;
(log z)−h .
The last part of the lemma follows from the inequality E/φ(E) ¿ log log E.
6 x.
Lemma 2.3. Let V (x) be the number of distinct values of φ(n) which are
Then for every ε > 0,
V (x) ¿ε
x
.
(log x)1−ε
Proof. This is the main theorem of [8].
Lemma 2.4. The number of n 6 x divisible by a number s satisfying
s > exp{(log log x)2 } and P + (s) 6 s10/ log log x is ¿ x/ log10 x.
Proof. For exp{(log log x)2 } 6 y 6 x, let G(y) be the number of s 6 y with
P + (s) 6 s10/ log log x . Let n denote a number with no prime factor exceeding
THE NUMBER OF SOLUTIONS OF φ(X)=M
W := y 10/ log log x , and set α = 1 −
G(y) 6
X ³ y ´α
n6y
n
<y
α
20 log log x
.
log y
∞
X
n=1
289
Then
Y
y
(1 − p−α )−1
(log x)20
p6W


X
y
¿
exp 
p−α  .
(log x)20
n−α =
p6W
By the prime number theorem,
Z W
X
dt
−α
p 6 (1 + o(1))
α
t log t
2
p6W
6 (1 + o(1))
Z
200
(log y)−1
eu du
u
6 (1 + o(1))(e200 + log log y) 6 2 log log x
for large x. Thus G(y) ¿ y/(log x)18 and the lemma follows by partial summation. For a survey of results on the distribution of numbers without large
prime factors, see [15].
The next four lemmas are needed to bound the product P (D, z) mentioned
in the introduction.
Lemma 2.5. When (a, n) = 1, n is odd and square-free,
¯
¶¯
µ
¯ X
aj + b ¯¯
¯
¯ 6 1.
¯
n
1 6j 6n
(j,n)=1
When (ab, n) = 1 and n is square-free,
¯
¯
¯ X µ ¶µ
¶¯
¯
¯
j
aj
+
b
¯
¯ = 1.
¯
¯
n
n
¯1 6j 6n
¯
Proof. Denote by f (n; a, b) the first sum in the lemma. The desired bound
follows from the identities f (n; a, b) = −( nb ) for prime n, and
f (n1 n2 ; a, b) = f (n1 ; an2 , b)f (n2 ; an1 , b)
when (a, n1 n2 ) = (n1 , n2 ) = 1. The second assertion follows in a similar
fashion. Here we adopt the convention that ( a1 ) = 1 for all a.
Lemma 2.6. For some absolute constant y0 , if y > y0 , 1 6 c 6 2, E
and D ∈ Z, then
µ ¶¶c
Yµ
X cω(n) µ(n) µ D ¶
1 D
.
1−
64
$ $
n
n
$ 6y
$ -E
n6y log log y
(n,2E)=1
P + (n)6y
> 1,
290
KEVIN FORD
Proof. If D = 0, the left side is 1 and the right side is 4, so suppose D 6= 0.
For 1 6 c 6 2 and x > −1/3, (1 + x)c 6 (1 + cx)(1 + 3x2 ); thus
Yµ
$ 6y
$ -E
1−
1
$
µ
D
$
¶¶c
6 94
Yµ
$ >3
1+
3
$2
¶ Y µ
3 6$ 6y
1−
c
$
µ
D
$
¶¶
$ -E
µ
¶
3
9 Y
1+ 2
=
4
$
$ >3
6 3.92
Since cω(n)
X
P + (n)6y
(n,2E)=1
n6y log log y
X
cω(n) µ(n)
n
P + (n)6y
(n,2E)=1
cω(n) µ(n)
n
µ
D
n
µ
¶
D
n
¶
X
+ 3.92
n>y log log y
P + (n)6y
cω(n)
.
n
6 τ (n), by Lemma 2.4 and partial summation, we obtain
X
n>y log log y
P + (n)6y
cω(n)
n
X
6
d1 d2 >y log log y
P + (d1 d2 )6y
1
¿
d1 d2
X
d2 >y (log log y)/2
P + (d2 )6y
1 X 1
d2
d1
d1 6d2
X
¿
d2 >y
P + (d2 )
(log log y)/2
log d2
d2
log log d2
6d3/
2
¿ (log y)−8 .
As the left side in the lemma is À (log y)−c , the desired bound follows for
sufficiently large y. In particular, the sum in the lemma is positive and À
(log y)−c .
Lemma 2.7. Suppose y > y0 , U > y 10(log log y) , A, B and R are positive
integers with R even and log(ABR) 6 log3 U . Suppose D(s) = A(A − Bs),
D(s) = As(As − B) or D(s) = A ± Bs. Uniformly in A, B, R and 1 6 c 6 2,
2
X Yµ
s 6U $ 6y
s prime
X
Yµ
$ 6y
s 6U
s,sR+1 prime
1
1−
$
1
1−
$
µ
µ
D(s)
$
D(s)
$
¶¶c
¿
U (log log U )1+c
,
log U
¿
U (log log U )4+c
.
(log U )2
¶¶c
291
THE NUMBER OF SOLUTIONS OF φ(X)=M
Proof. Let z = U 1/ log log U . If D(s) = A(A − Bs) or D(s) = As(As − B),
set E = AB; otherwise set E = B. By imposing the addition condition
that $ - E, we see that the products on the left are only increased when
D(s) = A(A − Bs) or D(s) = As(As − B), and in the case D(s) = A + Bs the
product is changed by a factor
µ ¶¶c
Yµ
1 A
1−
¿ (log log B)c ¿ (log log U )c .
$ $
$|B
We will also replace the conditions that “s prime” and “s, sR + 1 prime”,
with P − (s) > z, respectively P − (s(sR + 1)) > z. This will eliminate at
most z values of s from the sums on the left, reducing the sums by at most
O(z(log y)2 ) = Oε (U ε ). Let S1 denote the set of s 6 U with P − (s) > z and
let S2 denote the s 6 U with P − (s(sR + 1)) > z. By Lemma 2.6,
µ
¶¶c
X Yµ
X cω(n) µ(n) µ D(s) ¶
X
1 D(s)
1−
6
4
$
$
n
n
s∈S
s∈S
1
$ 6y
$ -E
1
64
n6y log log y
(n,2E)=1
P + (n)6y
X
n6y log log y
(n,2E)=1
¯
¶¯
µ
cω(n) µ2 (n) ¯¯ X D(s) ¯¯
¯
¯.
¯
¯
n
n
s∈S
1
Denote by S(S ; n, b) the number of elements of the set S which lie in the
residue class b mod n. By the sieve fundamental lemma (Lemma 2.1), and the
fact that n 6 z 1/5 , uniformly in (b, n) = 1 we have
S(S1 ; n, b) =
where
W1 =
Y
$ 6z
$ -n
¢
U W1 ¡
1 + O((log U )−10 ) ,
n
(1 − 1/$) ¿
n log log U
.
φ(n) log U
For each n, Lemma 2.5 gives
¶
µ
X µ D(s) ¶
X
D(b)
=
S(S1 ; n, b)
n
n
s∈S1
(b,n)=1
µ
¶
D(b)
U W1 X
=
+ O(U (log U )−10 )
n
n
(b,n)=1
U
U log log U 1
+
,
¿
log U
φ(n) log10 U
292
KEVIN FORD
and the first inequality follows. The second inequality follows in the same
manner by summing over s ∈ S2 and using
S(S2 ; n, b) =
U W2
(1 + O((log U )−10 )),
n
where W2 = 0 if (n, 1 + bR) > 1 and otherwise
W2 =
Y µ
$ 6z
$-nR
2
1−
$
¶
Y
$ 6z
$|R,$ -n
µ
1
1−
$
¶
µ
¿
nR
φ(nR)
¶2
(log log U )2
.
log2 U
Lemma 2.8. For some absolute constant b > 0, the following holds. For
each Q > 3, there is a number dQ ÀA (log Q)A for every A > 0, so that
whenever |D| 6 Q, dQ - 8D, and log z 6 exp{(log Q)b },
µ ¶¶
µ
Y
1 D
¿ 1.
1−
$ $
log log Q
Q
<$ 6z
Furthermore, 16 - dQ and p2 - dQ for all odd primes p.
Remark. From now on dQ will refer to the number in this lemma. The
implied constant in dQ ÀA (log Q)A is ineffective for A > 2. With more work,
the set of exceptional D may be reduced to those with f (D) = f (dQ ), where
f (n) is the product of the odd primes which divide n to an odd power. The
upper limit of z may also be extended easily.
Proof. We may assume without loss of generality that D is square-free.
If |D| 6 2 the result follows from the prime number theorem for arithmetic
D
) then depends only on $ modulo 8. Next suppose
progressions, since ( $
|D| > 3. The desired bound follows from a version of the prime number
theorem for arithmetic progressions deducible from the Landau-Page theorem
and a zero density result of Gallagher (see [6] and [12]). The number dQ
is the order of a real, primitive Dirichlet character whose corresponding Lfunction has a real zero β > 1 − c0 / log Q. If c0 > 0 is a sufficiently small
absolute constant, the Landau-Page theorem asserts that there is at most one
such primitive character of order 6 8Q. Furthermore, if this character exists,
Siegel’s Theorem implies that dQ ÀA (log Q)A for any A > 0. Let π(y; q, a)
denote the number of primes p 6 y, p ≡ a (mod q). For n 6 8Q not divisible
by dQ , (a, n) = 1 and y > Q0 := Qlog log Q , Theorem 7 of [12] implies
´
³
y
1 + O((log Q)−b )
π(y; n, a) =
φ(n) log y
THE NUMBER OF SOLUTIONS OF φ(X)=M
293
for some absolute constant b, 0 < b < 1. Therefore, for z > Q0 ,
¶
µ
X
1
log log z
log log z − log log Q0
.
=
+O
$
φ(n)
φ(n)(log Q)b
Q0 <$ 6z
$≡a (mod n)
Write D = D1 D2 , where D2 is the largest odd positive divisor of D (i.e.
D1 ∈ {−2, −1, 1, 2}). Since D2 > 3, quadratic reciprocity implies
µ ¶
µ ¶µ ¶
µ ¶
D
$
$
($−1)(D2 −1)/4 D1
= (−1)
= i$
,
$
$
D2
D2
where i$ ∈ {−1, 1} depends only on $ modulo 8. Therefore, for numbers
ie ∈ {−1, 1},
µ ¶
X 1 µD¶
X
X
X
a
1
=
ie
$ $
D2
$
Q0 <$ 6z
e=1,3,5,7
16a68D2
a≡e (mod 8)
Q0 <$ 6z
$≡a (mod 8D2 )
¶
µ
¶
D2 µ
8a0 + e
log log z
log log z − log log Q0 X X
+O
.
ie
=
φ(8D2 )
D2
(log Q)b
0
e
a =1
Here we have used the fact that ( Da2 ) = 0 if
by hypothesis, and the sum on a0 is zero,
(a, D2 ) > 1. The error term is O(1)
since as a0 runs from 1 to D2 , 8a0
runs over a complete system of residues modulo D2 . The lemma now follows
from the fact that 1 + h = exp{h + O(h2 )} whenever − 12 6 h 6 12 .
3. Proof of Theorem 2, Part I
Lemma 3.1. If A(m) = k and S(m) is true, then A(mm0 ) = k + 2 for
some m0 .
Proof. Suppose φ(α1 ) = · · · = φ(αk ) = m. Suppose x is sufficiently large,
so that x > 2αi for every i and there are À x/ log2 x primes q ∈ (x/2, x] such
that qm + 1 is prime. For each such q, the equation φ(n) = q(q − 1)m has at
least k + 2 trivial solutions, namely n = q 2 αi (i = 1, . . . , k), n = q(qm + 1)
and trivial n = 2q(qm + 1). Let B(x) denote the number of such q giving rise
to additional solutions. If φ(n) = q(q − 1)m, then clearly q 3 - n. If q 2 | n, then
φ(n/q 2 ) = m, which implies n is one of the trivial solutions. If q | n, q 2 - n,
then φ(n/q) = qm. There is a prime r | (n/q) with q | (r − 1), i.e. r = dq + 1
for some d | m. If d = m then n is one of the trivial solutions mpq + 1 or
2(mpq + 1). By Lemma 2.2, the number of q with dq + 1 for some other d is
O(x(log x)−3 ). The last case, counting solutions with (n, q) = 1, is codified in
a lemma, since we will need the argument in Part II. It follows from the next
294
KEVIN FORD
lemma that the number of q with φ(n) = q(q − 1)m for some n with (n, q) = 1
is O(x(log x)−2.1 ). Therefore, B(x) = O(x(log x)−2.1 ) and hence for some q,
A(q(q − 1)m) = k + 2.
Lemma 3.2. Suppose b = 1 or b is a prime less than y 10 . Let N (y) be
the number of primes q ∈ (y/2, y] satisfying (i) qbm + 1 is prime, and (ii) for
some n with (n, q) = 1, φ(n) = mq(q − 1). Then
N (y) ¿ y(log y)−2.1 .
Proof. First, we remove from consideration those q for which q − 1 has a
divisor s with s > exp{(log log y)2 } and P + (s) 6 s1/ log log y . By Lemma 2.4,
the number of such q is O(y/ log10 y). Consider the equation φ(n) = mq(q − 1).
There is a number r | n with q | r − 1. Say r = 1 + qs0 and let w0 = φ(n/r).
For some factorization of m as m = m1 m2 , we have m1 | s0 , m2 | w0 . Fix m1
and m2 , and let s = s0 /m1 , w = w0 /m2 , so that sw = q − 1. The number of q
in question is at most the number of pairs (s, w) for which y/2 − 1 6 sw 6 y,
wm2 is a totient (value taken by φ(n)), and the following three numbers are
prime:
(3.1) ws + 1 = q,
(ws + 1)mb + 1 = qmb + 1,
(ws + 1)sm1 + 1 = r.
For brevity write E(c) = exp{(log y)c }. For fixed s, Lemma 2.2 implies that
the number of w 6 y/s with each number in (3.1) prime is
¿
y (log log y)3
.
s log3 (y/s)
Therefore, the number of pairs (s, w) with s 6 E(0.89) is O(y(log y)−2.1 ).
Otherwise w 6 y/E(0.89). By Lemma 2.1 and (2.1), for fixed w 6 y/E(0.89),
the number of s with each number in (3.1) prime is
µ
¶¶
µ
1 m21 − 4m1 w
y(log log y)3 Y
,
1−
(3.2)
¿
w(log y)2.67 $6z
$
$
where z = E(0.89)1/2 .
First, suppose that w ∈ (U/2, U ] where U is a power of 2 less than 2E(1/2).
Let Q = 8mE(1/2) and recall the definition of dQ (Lemma 2.8). The number
of w with dQ | (m21 − 4m1 w) is clearly
6 4md 1 U
Q
+1¿
U
U
¿
.
2
(log Q)
log z
For the remaining w, Lemma 2.8 implies that
µ
¶¶
µ
Y
1 m21 − 4m1 w
¿ 1.
1−
$
$
log log Q
Q
6$ 6z
295
THE NUMBER OF SOLUTIONS OF φ(X)=M
Hence, since w is a totient, Lemma 2.3 gives
µ
¶¶
µ
X
1 Y
(log y)1/2 log log y
1 m21 − 4m1 w
.
¿
1−
w $ 6z
$
$
(log U )0.999
U/2<w6U
By (3.2), summing on U gives a total of O(y(log y)−2.16 ) possible pairs (s, w).
Lastly, consider E(1/2) 6 w 6 y/E(0.89). The product in (3.2) for
$ > E(0.48) is O((log y)0.41 ). By our initial assumption about q, w has a
prime factor > E(0.49). Thus n/r has a prime divisor r2 > E(0.49). We can
assume that r22 - n, for the number of q divisible by a factor r2 (r2 − 1) for such
large r2 if O(y/E(0.49)). Write w∗ = φ(n/(rr2 )), so that w = (r2 − 1)w∗ . Fix
w∗ and put r2 ∈ (U/2, U ], where U is a power of 2 in [E(0.49), 2y/(sw∗ )]. By
Lemma 2.7,
µ
¶¶
X
Y µ
1
(log log y)2
1 m21 − 4m1 w∗ (r2 − 1)
¿
.
1−
r2
$
$
log U
U/2<r2 6U
$6E(0.48)
Summing on U and again using (3.2) and Lemma 2.3, we see that the total
number of pairs (s, w) is O(y(log y)−2.25 ). Thus, for fixed m1 , m2 , the number
of possible q is O(y(log y)−2.1 ). Since the number of choices for (m1 , m2 ) is
O(1), the lemma follows.
4. Part II, the case (n, pq) > 1
We require one more sieve lemma, needed only for a special case of the
proof for Part II.
Lemma 4.1. Suppose r, s, t, u are positive integers, and x, y, z are positive
real numbers satisfying x > y > z > 2 and log(rstu) 6 log3 x. Let
wg,h = (rg + 1)(sh + 1)
and
F (g, h) = (twg,h + 1)(ughwg,h + 1)ghwg,h .
Then
6
#{x < g
log x)
6 2x, y < h 6 2y : P − (F (g, h)) > z} ¿ xy (log
(log z)6
.
Proof. Let %($) denote the number of solutions of F (g, h) ≡ 0 (mod $).
If %($) = $2 for some $ 6 z then the left side is zero. Otherwise by the sieve
fundamental lemma (Theorem 2.5 of [13]),
¶
Yµ
%($)
−
#{x < g 6 2x, y < h 6 2y : P (F (g, h)) > z} ¿ xy
1−
.
$2
$ 6z
296
KEVIN FORD
We next show that for $ - 30rstu,
−12 − 2$3/4
(4.1)
6 %($) − 6$ 6 −10 + 2$3/4 .
From (4.1), the lemma follows at once, since then
¶6 Y
¶
µ
Yµ
30rstu
%($)
(1 − 1/$)6
1−
¿
2
$
φ(30rstu)
$ 6z
$ 6z
¿
(log log x)6
.
(log z)6
To prove (4.1), suppose $ 6 z, $ - 30rstu. Let M1 , M2 , M3 , M4 denote the set
of solutions mod $ of gh ≡ 0, wg,h ≡ 0, twg,h + 1 ≡ 0 and ughwg,h + 1 ≡ 0,
respectively. It is easily verified that M1 ∩ M4 = M2 ∩ M3 = M2 ∩ M4 = ∅,
|M1 ∩ M2 | = |M1 ∩ M3 | = 2 and |M3 ∩ M4 | 6 2. Also, |M1 | = |M2 | = 2$ − 1
and |M3 | = $ − 1. By inclusion-exclusion,
%($) = |M1 ∪ M2 ∪ M3 ∪ M4 | = 5$ − 7 + |M4 | − |M3 ∩ M4 |.
Since the number of solutions of q(qR + 1) ≡ a (mod $) is 1 + ( 1+4aR
$ ), we
have
¶¶ µ
µ
¶¶
µ
$−1
Xµ
1 − 4sa−1 u−1
1 + 4ar
1+
1+
|M4 | =
$
$
a=1
$−1
X µ 1 + 4ar ¶ ³ a ´ µ a − 4su−1 ¶
=$−3+
=: $ − 3 + E.
$
$
$
a=1
By Theorem 1 of Davenport [5], we have |E| 6 2$3/4 and (4.1) follows.
We next combine Lemmas 2.7 and 2.8 into an estimate which has enough
generality for our applications.
Lemma 4.2. Suppose δ > 0, z > 100. Suppose A, B, Q, R, U are positive
integers satisfying (log z)δ 6 log Q 6 (log z)3 , log(ABRU ) < log3 z, U > 10,
1/2
(AB, dQ ) 6 dQ and Q > AU (AU + B). Suppose either D(s) = A(A − sB) or
D(s) = sA(sA − B). Let S1 denote the set of primes s 6 U , and S2 the set of
primes s 6 U with sR + 1 prime. If 1 6 c 6 2 and k ∈ {1, 2}, then
µ
µ
¶¶c
¶
X Yµ
1 D(s)
log Q c
U
1−
¿δ
(log log z)4+4c + log5 z.
k
$
$
log
U
(log
U
)
s∈S $6z
k
Proof. Partition Si into S 0 , the set of s ∈ Si for which dQ - AD(s), and
S 00 , the remaining s. If D(s) = A(A − sB), then S 00 is contained in a single
1/2
residue class modulo g, where g = dQ /(dQ , AB) > dQ . Recall that 16 - dQ
297
THE NUMBER OF SOLUTIONS OF φ(X)=M
and dQ is not divisible by the square of any odd prime. If D(s) = sA(sA − B),
then (s, dQ ) > 1 for at most ω(dQ ) 6 log Q 6 log3 z primes s (Here we use
dQ 6 8Q). For other s, dQ |A(sA − B) for s lying in a single residue class
modulo g = d0 /(d0 /, A), where d0 = dQ /(dQ , A). Here (d0 , A) 6 8, therefore
1/2
g > 18 dQ . Thus, in every case
|S 00 | ¿
U
U
+ log3 z,
+ 1 + log3 z ¿
g
(log Q)8/δ
so the contribution to the sum in the lemma from s ∈ S 00 is
O(U (log z)−6 + log5 z) = O(U/ log2 U + log5 z).
By Lemma 2.8, if s ∈ S 0 and z > Q0 := Qlog log Q then
µ
¶¶c
Y µ
1 D(s)
¿δ 1.
1−
$
$
Q0 <$ 6z
2
Set y = U 1/(20(log log U ) ) . Trivially we have
µ
µ
µ
¶¶c
¶
¶
Y µ
log Q0 c
log Q c
1 D(s)
¿
¿
(log log z)3c .
1−
$
$
log y
log U
y<$ 6Q0
Therefore, for any admissible z, Lemma 2.7 implies
µ
¶¶c
X Yµ
1 D(s)
1−
$
$
s∈S 0 $6z
¶
µ
¶¶c
µ
µ
X
Y
1 D(s)
log Q c
1−
¿
(log log z)3c
log U
$
$
s∈S 0 $6min(y,z)
µ
¶
log Q c
U
¿
(log log z)4+4c .
(log U )k log U
As mentioned in the introduction, for most primes p 6 x, the prime factors
of p − 1 are nicely distributed. We prove a simple result of this kind below.
For brevity write E(z) = exp{(log x)z }.
Lemma 4.3. Let N > 10 be an integer, put δ = 1/N and suppose x1/10
y 6 x. The number of primes p 6 y not satisfying
Ω(p − 1, E(jδ), E((j + 1)δ)) 6 (δ + δ 2 ) log log x
is
¿δ y(log x)−1−δ
3 /3
(0 6 j
6
6 N − 1)
.
Proof. Assume that x is sufficiently large with respect to δ. First, the
number of p 6 y for which P + (p − 1) 6 y 1/ log log x is O(y/ log10 x) by Lemma
298
KEVIN FORD
2.4. For the remaining p, let q = P + (p − 1) and p − 1 = qa. For fixed a, the
number of q 6 y/a with qa + 1 prime is, by Lemma 2.2,
¿
y log log x
y(log log x)3
¿
.
a log2 (y/a)
a log2 x
For each j, let Ij = [E(jδ), E((j + 1)δ)] and let Bj be the sum of 1/a over
a 6 y with at least (δ + δ 2 ) log log x − 1 prime factors in Ij . We have
Bj
6 (1 + δ)1−(δ+δ ) log log x
2
6 2(log x)−(δ+δ ) log(1+δ)
X (1 + δ)Ω(a,E(jδ),E(jδ+δ))
2
¿ (log x)1+δ
a
µ
¶
¶
Yµ
Y
p
1 + δ −1
1−
p−1
p
a 6x
p6x
p6∈Ij
2 −(δ+δ 2 ) log(1+δ)
p∈Ij
3
¿ (log x)1−0.4δ ,
and the lemma follows.
Suppose that Hypothesis S(m) (following Lemma 1.2) is false. By Lemma
1.2, for an unbounded set of x, there are À x/ log2 x primes s ∈ (x/2, x] for
which s = 1 + pqm, where p and q are primes and x1/10 6 p 6 x1/2 . Denote
by A the set of such pairs of primes (p, q). We will show that for x sufficiently
large, there is some such pair (p, q) giving A(mpq(p − 1)(q − 1)) = k + 2, with
(4.2)
φ(n) = mpq(p − 1)(q − 1)
having only the trivial solutions n = p2 q 2 αi (i = 1, . . . , k), n = pq(mpq + 1)
and n = 2pq(mpq + 1). Before proceeding, we must cull from A certain pairs
(p, q) possessing undesirable properties. Let F (p) be the number of q with
(p, q) ∈ A and let G(q) be the number of p such that (p, q) ∈ A . By Lemma
2.2, we have
(4.3)
F (p) ¿
x
,
p log2 x
G(q) ¿
x
.
q log2 x
By (4.3) and the Brun-Titchmarsh inequality, if K is a large enough constant
(depending only on m),
#{(p, q) ∈ A : p ≡ 1(mod 2K ) or q ≡ 1(mod 2K )} 6 12 |A |.
Denote by A 0 the set of remaining pairs. Define ν by 2ν | m, 2ν+1 - m. If n
has h distinct prime factors, then 2h−1 | φ(n). Consequently, if (4.2) holds and
(p, q) ∈ A 0 , ω(n) 6 C where C = 2K + 1 + ν. Let
(4.4)
N = 100([4 log(2C + 2)] + 14 + 2C),
δ = 1/N.
THE NUMBER OF SOLUTIONS OF φ(X)=M
For brevity, write Ej = exp{(log x)1−jδ } for 0
Define
3δ log log x
Qj = Ej−1
,
(4.5)
Let
B be the set of (p, q) ∈ A
0
6 j 6 N − 1 and let EN
dj = dQj ,
299
= 2.
(j = 2, . . . , N ).
satisfying the additional conditions
(4.6)
d2 | (p − 1)(q − 1) implies d 6 log4 x,
(4.7)
dp + 1, dq + 1, dpq + 1 are composite for each d | m,
except for mpq + 1,
(4.8)
p(p − 1) + 1, q(q − 1) + 1, pq(p − 1) + 1, pq(q − 1) + 1,
are all composite,
(4.9)
(m(p − 1)(q − 1), dj ) 6 dj
(4.10)
p - (q − 1),
and
(
(4.11)
1/2
(j = 2, . . . , N ),
Ω(p − 1, Ej+1 , Ej ) 6 (δ + δ 2 ) log log x
(0 6 j
6 N − 1)
(0 6 j 6 N − 1).
Ω(q − 1, Ej+1 , Ej ) 6 (δ + δ ) log log x
2
First, the number of (p, q) failing (4.6) does not exceed
X τ (d2 )
X X X ax
¿
x
log
x
¿ x(log x)−2.9 .
2
2
nd
d
4
2
4
d>log x a|d n<x,a|n
d>log x
Since m is fixed, by Lemma 2.2 and (4.3), the number of pairs failing (4.7) is
O(x/ log3 x). For (4.8), given q the number of p with pq(q − 1) + 1 prime is
O(x(log log x)3 /(q log3 x)) by Lemma 2.2. Thus the number of pairs (p, q) with
pq(q − 1) + 1 prime is O(x(log x)−2.99 ). Likewise the number of pairs satisfying
pq(p − 1) + 1 prime is O(x(log x)−2.99 ). Also, by Lemma 2.2 and the prime
number theorem for arithmetic progressions, for each q the number of p with
p(p − 1) + 1 prime is
µ ¶¶
Y µ
1 −3
x
x
1−
¿
.
¿
3
p
p
q log x
q log3 x
1/20
p6x
Thus the number of pairs (p, q) with p(p − 1) + 1 prime is O(x/ log3 x), and the
same bound holds for the number of pairs with q(q −1)+1 prime. To deal with
1/4
(4.9), for each j let Vj denote the set of numbers a 6 dj with (a, dj ) > dj /m.
Then
X
X
|Vj | =
φ(n/d) 6
n/d
1/4
d|dj ,d>dj
=
X
/m
1/4
d|dj ,d>dj
/m
3/4
4/5
d < τ (dj )mdj
3/4
d|dj ,d<mdj
¿ dj .
300
KEVIN FORD
By Lemma 2.8, dj À (log Qj )40/δ > (log x)40 and for x large, m 6 dj
for
every j. Therefore, the number of pairs (p, q) failing (4.9) is at most
¶
µ
¶
N
−1 X µ
N
−1
X
X
X
X
x
x
x log x
¿
|Vj |
+1
+
p
q
dj
1/100
j=1 a∈Vj
x1/10 <p6x1/2
p≡1+a (mod dj )
x1/2 <q 6x9/10
q≡1+a (mod dj )
j=1
¿
x
.
(log x)3
The number of pairs failing (4.10) is trivially O(x9/10 ). By (4.3), Lemma 4.3
and partial summation, the number of pairs failing (4.11) is at most
X
X
x
F (p) +
G(q) ¿
.
(log x)2+δ3 /3
bad p
bad q
Thus, for x large,
(4.12)
1
|B | > |A 0 | À x/ log2 x.
2
We will show that for most pairs (p, q) ∈ B , the equation (4.2) has only the
trivial solutions. We break this argument into two parts, the second of which
is the most delicate and is proved in the next section.
Lemma 4.4.
The number of pairs (p, q) ∈ B for which (4.2) has a
nontrivial solution n with (n, pq) > 1 is O(x(log x)−2.1 ).
Proof. Consider a generic solution (n, p, q) of (4.2) with (n, pq) > 1. By
the definition of B , p3 - n and q 3 - n. If (n, p2 q 2 ) = p2 q 2 , then φ(n/p2 q 2 ) =
m, which has the k trivial solutions n = p2 q 2 αi (i = 1, . . . , k). Next, if
(n, p2 q 2 ) = p2 q, then φ(n/p2 q) = mq. This implies that there is some prime
r | n with q|(r − 1)|mq, which is impossible by (4.7). Similarly, (n, p2 q 2 ) = pq 2
is impossible. If (n, p2 q 2 ) = pq, then φ(n/pq) = mpq, implying n has a prime
divisor of the form dp + 1, dq + 1 or dpq + 1 for some d | m. By (4.7),
n = pq(mpq + 1) or n = 2pq(mpq + 1). If (n, p2 q 2 ) = p2 , then φ(n/p2 ) =
mq(q − 1). By Lemma 3.2 (with b = p), for fixed p the number of possible
q is O((x/p)(log x)−2.1 ); therefore the total number of possible pairs (p, q) is
O(x(log x)−2.1 ). The same argument handles the case (n, p2 q 2 ) = q 2 .
Finally, suppose (n, p2 q 2 ) = p (the symmetric case (n, p2 q 2 ) = q is dealt
with the same way). Then φ(n0 ) = mpq(q − 1), where n0 = n/p, and there are
two cases: (i) there is a prime r | n0 with pq | (r − 1); (ii) there are primes
r1 , r2 dividing n0 with p | (r1 − 1) and q | (r2 − 1).
In case (i) write r = 1 + pqs0 and w0 = φ(n0 /r), so s0 w0 = m(q − 1). Fix
m1 and m2 , where m = m1 m2 , m1 | s0 and m2 | w0 . Write w = w0 /m2 and
s = s0 /m1 , so that sw = q − 1. For fixed s 6 E(3/4) and fixed p, the number
THE NUMBER OF SOLUTIONS OF φ(X)=M
301
of w with sw + 1 = q, psm1 (sw + 1) + 1 = r and pm(sw + 1) + 1 all prime is
¿
x (log log x)3
.
sp log3 x
Summing on s 6 E(3/4) and p yields a total of O(x(log x)−2.24 ) possible pairs
(p, q). The other possibility is that w < q/E(3/4). By Lemma 2.1 (with
z = E(0.74)) and (2.1), for fixed w and p, the number of s is
µ
¶¶
Y µ
1 m21 p2 − 4m1 pw
x (log log x)3
.
1−
¿
pw (log x)2.22
$
$
$6E(0.74)
Put p ∈ (U/2, U ], where U is a power of 2 between x1/10 and x1/2 . By Lemma
2.7, the above product has average O((log log x)2 ) over such p. Summing over
U , it follows that for fixed w, the number of pairs (p, s) is O((x/w)(log x)−2.21 ).
Since wm2 is a totient, Lemma 2.3 implies that the number of pairs (p, q) is
O(x(log x)−2.2 ). Therefore there are O(x(log x)−2.2 ) pairs (p, q) counted in case
(i), since there are O(1) choices for m1 and m2 .
In case (ii), write
r1 = 1 + ps1 ,
r2 = 1 + qs2 ,
n0 = r1 r2 t,
w0 = φ(t).
Fix m1 , m2 , m3 , where m = m1 m2 m3 , m1 | s1 , m2 | s2 and m3 | w0 . Let
s3 = s1 /m1 , s4 = s2 /m2 and w = w0 /m3 . The number of possible pairs (p, q)
is at most the number of quadruples (s3 , s4 , w, p) where p is prime, wm3 is a
totient, and the following four numbers are prime:
ws3 s4 + 1 = q,
(4.13)
ps3 m1 + 1 = r1 ,
(ws3 s4 + 1)s4 m2 + 1 = r2 ,
(ws3 s4 + 1)pm + 1 = pqm + 1.
Suppose U is a power of 2, x1/10 6 U 6 2x1/2 and U/2 < p 6 U . We will
count separately the quadruples (s3 , s4 , w, p) in each of three classes: 1) those
with w > E(0.99); 2) those with w < E(0.99), s > E(0.8); 3) those with
w < E(0.99) and s 6 E(0.8).
First, suppose w > E(0.99). By Lemma 2.2, given (s3 , s4 ), the number of
p with ps3 m1 + 1 prime is O(U (log log x)2 / log2 x) and for each p the number
of w with the remaining three numbers in (4.13) all prime is
x(log log x)3
x(log log x)3
¿
.
U s3 s4 (log x)2.97
U s3 s4 log3 (x/(U s3 s4 ))
By (4.11), Ω(s3 s4 ) 6 1.01 log log x, so
X 1
X
τ (m)
6
¿ (log x)1+1.01 log 2 .
s
s
m
3
4
s ,s
¿
3
4
m6x
Ω(m)61.01 log log x
302
KEVIN FORD
Therefore, the number of quadruples in the first class is O(x(log x)−2.2 ), since
there are O(log x) possibilities for U .
Secondly, suppose w < E(0.99) and s3 > E(0.8). Applying Lemma 2.2
twice, for each pair (s4 , w) the number of possible (p, s3 ) with p and the numbers in (4.13) all prime is
¿
(log log x)4
x
x(log log x)4
.
¿
ws4 (log3 x) log2 (x/(U s4 w))
ws4 (log x)4.6
Since wm3 is a totient, by Lemma 2.3 and partial summation we have
X1X 1
¿ (log x)1.01 .
w
s
4
w
s
4
Summing over U gives O(x(log x)−2.58 ) quadruples in the second class.
Finally, suppose s3 < E(0.8) and w < E(0.99). By Lemma 2.2, for each
pair (s3 , w), the number of (p, s4 ) with p, ps3 m1 + 1 = r1 , s3 s4 w + 1 = q and
(s3 s4 w + 1)pm + 1 all prime is
¿
x(log log x)4
x(log log x)4
¿
.
s3 w log2 x log2 (x/(U s3 w))
s3 w log4 x
We have ignored the fact that r2 is prime, because r2 is a quadratic function of
s4 . As in the previous case, wm3 being a totient implies that the sum of 1/w is
¿ (log x)0.01 . Since the sum of 1/s3 is O((log x)0.8 ), there are O(x(log x)−2.18 )
quadruples in the third class. The number of triples (m1 , m2 , m3 ) is O(1), so
there are O(x(log x)−2.18 ) pairs (p, q) counted in case (ii).
5. Part II, the case (n, pq) = 1
In this argument we use for the first time the fact that the prime factors
of p − 1 and q − 1 are nicely distributed (i.e. the full strength of (4.11)). Recall
that Ej = exp{(log x)1−jδ } for 0 6 j 6 N − 1 and EN = 2.
We may assume that any prime divisor of n which is > EN −1 divides n
to the first power, for the number of (p, q) with r(r − 1) | m(p − 1)(q − 1) for
some r > EN −1 is O(x/EN −1 ). Denote by B 0 the set of remaining pairs (p, q).
Let r1 , . . . , rM be the prime factors of n satisfying
P + (ri∗ ) > EN −1 ,
Clearly 1 6 M
6 C.
ri∗ :=
ri − 1
.
(ri − 1, pq)
Let w = φ(n/(r1 · · · rM )), so that
w
P + (w∗ ) 6 EN −1 ,
w∗ =
.
(w, pq)
303
THE NUMBER OF SOLUTIONS OF φ(X)=M
In our new notation, we have
∗
w∗ = m(p − 1)(q − 1).
r1∗ · · · rM
(5.1)
We will partition the solutions of (5.1) according to the value of M (the number
of “large” prime factors of n), the size of the largest prime factor of each ri∗ ,
and the location of the factors p and q among the factors ri − 1 and w. For
convenience, set r0 = w+1. Define numbers i0 and i00 by p | (ri0 −1), q | (ri00 −1).
It is possible that i0 = i00 . Let Ij = [Ej , Ej−1 ) and define numbers ji by
P + (ri∗ ) ∈ Iji . The primes ri may be ordered so that 1 = j1 6 j2 6 · · · 6 jM .
The fact that j1 = 1 follows from the j = 0 case of (4.11).
By (4.8), it follows that p − 1 6= ri∗0 and q − 1 6= ri∗00 . Thus we may define
µ
¶
µ
¶
(p − 1)ri∗0
(q − 1)ri∗00
0
+
00
+
(5.2)
S =P
,
S =P
.
(p − 1, ri∗0 )2
(q − 1, ri∗00 )2
In other words, S 0 is the largest prime dividing p−1 and ri∗0 to different powers.
Define j 0 , j 00 by S 0 ∈ Ij 0 , S 00 ∈ Ij 00 , so that j 0 and j 00 measure the rough sizes of
S 0 and S 00 . By (4.11), if j 0 > 1 then the prime factors of p − 1 which are > E1
also divide ri∗0 , so that ji0 = 1. Similarly, if j 00 > 1 then ji00 = 1. Furthermore,
if either j 0 > 1 or j 00 > 1 then there are at least two values of i with ji = 1.
Thus, without loss of generality, if j 0 > 1 we may assume i0 = 1 and if j 00 > 1,
we may assume i00 = 2.
We count separately the number of (p, q) ∈ B 0 giving solutions of (5.1)
with M , j1 , . . . , jM , i0 , i00 , j 0 and j 00 fixed. The number of choices for these
parameters is O(1) since it only depends on m.
To reap the full benefit of (4.11), we break each factor in (5.1) into N
pieces as follows:
m(p − 1) =
N
Y
πj ,
q−1=
j=1
N
Y
θj ,
ri∗
=
j=1
N
Y
%i,j
(1 6 i 6 M ),
j=1
where πj , θj and %i,j are composed only of primes in Ij . We can suppose x
is large enough so that P + (m) < EN −1 , i.e. m | πN . Some of these variables
may equal 1. In particular, for each i we have %i,j = 1 for j < ji . By the
definition of j 0 and j 00 we also have
(5.3)
πj = %1,j
(j < j 0 ),
θj = %2,j
(j < j 00 ).
Let
(5.4)
aj = |{i > 1 : ji
6 j}|
(j = 1, . . . , N − 1),
b0 = |{i > 1 : (ri − 1, pq) > 1}|,
bj = |{i > 1 : ji = j, (ri − 1, pq) = 1}|,
(j = 1, . . . , N − 1).
304
KEVIN FORD
For each j, aj is the maximum number of %i,j which can be > 1. The estimation
of the number of solutions of (5.1) proceeds in N steps; at step j we count the
possibilities for πj , θj and the %i,j , assuming the variables consisting of smaller
primes have been fixed. The possibility that j 0 > 1 or j 00 > 1 will force us to
express some bounds in terms of products involving the Legendre symbol, and
these bounds must be carried over many steps. For convenience, for each j > 1
set
Y
Y
Y
(5.5)
Ri,j =
%i,h , Pj =
πh , Tj =
θh .
h>j
If
j0
h >j
h >j
> 1 define
(5.6)
D1 = R1,j 0 (R1,j 0 − 4Pj 0 /m),
H1 =
and if j 00 > 1 define
(5.7)
D2 = R2,j 00 (R2,j 00 − 4Tj 00 ),
H2 =
Note that by (5.3),
µ ¶ µ ∗ ∗
¶
D1
r1 (r1 − 4(p − 1))
(5.8)
=
,
$
$
µ
Y µ
$6E1
Y µ
$6E1
D2
$
¶
1
1−
$
µ
=
1
1−
$
µ
µ
D1
$
D2
$
¶¶
,
¶¶
.
r2∗ (r2∗ − 4(q − 1))
$
¶
.
Below are the estimates that we will prove:
Step 1. Let w∗ , P2 , T2 and R1,2 , . . . , RM,2 be given. The number of
possibilities for π1 , θ1 and %1,1 , . . . , %M,1 is
x
¿
(log log x)3+b0 +b1 +2a1 (log x)−4−b0 −b1 +δ(3+b0 +b1 +2a1 ) X1 X2 ,
P2 T2
½
where
X1 =
Next, define
Yj =
½
1
j0 = 1
H1
j0 > 1
1
j 6= j 0
H1
j = j0
½
,
X2 =
½
,
Zj =
1
j 00 = 1
H2
j 00 > 1
1
j 6= j 00
H2
j = j 00
.
.
Step j, 2 6 j 6 N − 1. Let w∗ , Pj+1 , Tj+1 and R1,j+1 , . . . , RM,j+1 be
given. Then
X Yj Zj
¿ (log log x)2bj +14ζj (log x)2δ(1+δ) log aj +2δ+bj (jδ−1)+4ζj δ ,
πj θj
where the sum is over all possible πj , θj and %1,j , . . . , %M,j , and ζj = 1 if j = j 0
or j = j 00 and ζj = 0 otherwise.
305
THE NUMBER OF SOLUTIONS OF φ(X)=M
Step N . We have
X YN ZN
¿ (log log x)14 (log x)5δ+2δ(1+δ) log(2M +2) ,
πN θN
where the sum is over w∗ , πN , θN , and %i,N (1 6 i 6 M ).
Using the fact that b0 + · · · + bN −1 = M and combining all N steps, the
total number of solutions of (5.1) is
¿ x(log log x)2M +45+a1 (log x)Y ,
where

Y = −4 − M + 2δ 2 
N
−1
X

log aj + log(2M + 2) + δX,
j=2
X = 16 + b0 + b1 + 2a1 +
N
−1
X
(jbj + 2 + 2 log aj ) + 2 log(2M + 2).
j=2
Since aj 6 M for each j, the coefficient of δ 2 is at most 2N log(2M + 2). Also,
b0 6 2; thus
X
6 2 log(2M + 2) + 14 + 2N + 2M +
N
−1
X
(2 log aj + jbj ).
j=1
The number of aj = i is ji+1 − ji . Also, j1 = 1 and 2 log 2 > 1; thus, if M > 2
then
N
−1
M
M
−1
X
X
X
(2 log aj + jbj ) 6
ji + 2
(ji+1 − ji ) log i + 2(N − 1 − jM ) log M
j=1
i=1
i=1
= 2(N − 1) log M + j1 +
M ³
X
1 − 2 log
³
i
i−1
´´
ji
i=2
6 2N log M + N
M ³
X
1 − 2 log
³
i
i−1
´´
i=3
= N (M − 2 + 2 log 2).
The above bound also holds when M = 1, since ai = 1 for all i and bj = 0 for
j > 1. Therefore, by (4.4), we obtain
Y
6 −4 + 2 log 2 + δ(14 + 2M + 4 log(2M + 2)) 6 −2.603
for every possible M , i0 , i00 , j 0 , j 00 and j1 , . . . , jM . In conclusion, the number
of solutions of (4.2) with (n, pq) = 1 is O(x(log x)−2.6 ). Together with (4.12)
and Lemma 4.4, this completes the proof of Part II of Theorem 2.
We now establish the bounds claimed for each of the N steps.
306
KEVIN FORD
Proof of Step 1 bound. Let I = {i : ji = 1} = {1, 2, . . . , a1 }. Recall (5.5)
and for i ∈ I write
%i,1 = Ai Bi ,
(5.9)
Ai = (%i,1 , π1 ),
Then
π1 =
(5.10)
Y
Ai ,
θ1 =
i∈I
and
P−
(5.11)
Bi = (%i,1 , θ1 ).
Y
Bi
i∈I
µY
¶
Ai Bi
> E1 .
i∈I
We need to count the number of (A1 , . . . , Ba1 ) satisfying (5.11) such that the
following 3 + b0 + b1 numbers are prime:










(5.12)









π1 P1 /m + 1 = p,
θ1 T1 + 1 = q,
(π1 P1 /m + 1)(θ1 T1 + 1)m + 1 = pqm + 1,
Ai Bi R1,i (ri − 1, pq) + 1 = ri ,
R1,i (ri − 1, pq) + 1 = ri ,
(i ∈ I),
(i = i0 6∈ I or i = i00 6∈ I).
Recall that p | (ri0 − 1) and q | (ri00 − 1). All variables Ai , Bi occur linearly on
the left of the above expressions except in a few cases. If i0 ∈ I (i.e. ji0 = 1),
then Ai0 occurs quadratically in the expression for ri0 (fourth line in (5.12)),
and if i00 ∈ I then Bi00 occurs quadratically in the expression for ri00 (fourth line
of (5.12)). There are O(1) possibilities for I1 = {i : Ai > 1}, I2 = {i : Bi > 1}.
Fix I1 , I2 and note that both I1 and I2 are nonempty and I1 ∪ I2 = I. Let
J = |I1 | + |I2 |.
Place each variable which is > 1 into a dyadic interval, e.g. Ui < Ai 6 2Ui
for i ∈ I1 , Vi < Bi 6 2Vi for i ∈ I2 , where each Ui , Vi is a power of 2. For
some ordering of the J variables, we use a sieve to bound the possibilities
for the first variable with the other variables undetermined. Then with the
first variable fixed, we use the sieve to bound the possibilities for the second
variable, with the remaining J − 2 variables undetermined, and so on. In most
cases the variables may be ordered so as to apply Lemma 2.2 with each. The
exceptions are when either Ai0 appears quadratically in the expression for ri0
and is isolated (none of the other J − 1 variables appears in the expression for
ri0 ), or when Bi00 has a similar property, or when i0 = i00 = 1 and Ai0 and Bi00
are isolated in the expression for ri0 . We therefore distinguish three cases: (i)
a1 = i0 = i00 = 1; (ii) j 0 = j 00 = 1 other than case (i); (iii) j 0 + j 00 > 2.
THE NUMBER OF SOLUTIONS OF φ(X)=M
307
In case (i), we have J = 2, j 0 = j 00 = 1, b0 = 1, b1 = 0, π1 = A1 and
θ1 = B1 . By Lemma 4.1, the number of pairs A1 , B1 satisfying (5.11) such
that p = A1 (P1 /m) + 1, q = B1 T1 + 1, pqm + 1 and A1 B1 R1,1 pq + 1 = r1 are
prime is O(U1 V1 (log log x)6 (log E1 )−6 ).
In case (ii), we first choose the numbers Ai0 (if i0 > 0) and Bi00 (if i00 > 0).
Since j 0 = j 00 = 1, none of the primes in (5.12) are determined solely by these
choices, so no sieving is required. The remaining variables Ai and Bi occur
linearly on the left of the expressions in (5.12), so we may apply Lemma 2.2
repeatedly. The number of solutions counted is
µY
Y ¶
Ui
Vi (log log x)3+b0 +b1 +J (log E1 )−(3+b0 +b1 +J) .
¿
i∈I1
i∈I2
For case (iii), suppose j 0 > 1, i0 = 1, so that π1 = %1,1 = A1 and r1 =
A1 R1,1 (A1 P1 /m + 1) + 1. By Lemma 2.1 and (5.8), the number of possible A1
is ¿ U1 H1 (log log x)2 (log E1 )−2 . Similarly, if j 00 > 1 the number of possible
B2 is ¿ U2 H2 (log log x)2 (log E1 )−2 . Lemma 2.2 is applied repeatedly to the
other variables yielding a total of
µY
Y ¶
¿
Ui
Vi (log log x)3+b0 +b1 +J (log E1 )−(3+b0 +b1 +J) X1 X2
i∈I1
i∈I2
solutions counted.
In every case, with I1 and I2 fixed,
Y
X Y
Ui
Vi ¿
Ui ,Vi i∈I1
i∈I2
x
(log x)J−1
P2 T 2
and the bound in Step 1 follows.
Proof of bound for step j ∈ [2, N − 1]. All j. There are bj numbers i with
ji = j and (ri − 1, pq) = 1. Recall (5.2) and let I denote the set of such i
with the additional property that S 0 - %i,j and S 00 - %i,j . For each i ∈ I, let
Wi = P + (%i,j ). Fix the partition of these i into I1 , the set of i with Wi | θj ,
and I2 , the set of i with Wi | πj . There are O(1) such partitions. Also fix
τ := {%∗1,j , . . . , %∗M,j ; πj∗ , θj∗ }.
Here, if j 6= j 0 and j 6= j 00 , we take %∗i,j = %i,j /Wi if i ∈ I and %∗i,j = %i,j
Q
Q
otherwise, πj∗ = πj / i∈I2 Wi and θj∗ = θj / i∈I1 Wi . If j = j 0 , we also divide
out the factor S 0 from either πj or θj , as well as the appropriate %i,j . If j = j 00
we divide out the factor S 00 in the same manner. Let
(5.13)
t = %∗1,j · · · %∗M,j = πj∗ θj∗
and note that by (4.11),
(5.14)
Ω(t) 6 2δ(1 + δ) log log x.
308
KEVIN FORD
Our procedure will be to fix τ and the Wi (i ∈ I), and sum Yj Zj /S 0 S 00 over
S 0 , S 00 if necessary. Then for each i ∈ I, Wi and Wi %∗i,j Ri,j+1 + 1 are prime, so
by Lemma 2.2 and partial summation,
X 1
(5.15)
¿ (log log x)2 (log x)jδ−1 .
Wi
Wi
For fixed t, let h(t) be the number of (M + 2)-tuples τ with product t. At
most aj of the numbers %i,j are > 1, thus h(t) is at most the number of dual
factorizations of t into 2 and aj factors, respectively. Actually if j < max(j 0 , j 00 )
we can do better, but this bound suffices. Therefore h(t) 6 (2aj )Ω(t) . Since
2 6∈ Ij , by (5.14) we have
X h(t)
(5.16)
t
t
log log x
6 a2δ(1+δ)
j
X 2Ω(t)
t
6 (log x)2δ(1+δ) log a
j
t
Y
(1 − 2/p)−1
p∈Ij
¿ (log x)
2δ(1+δ) log aj +2δ
.
All that remains is to take care of the sum on S 0 , S 00 .
Proof of bound for step j ∈ [2, N − 1]; j 6= j 0 , j 6= j 00 . Here there is no
sum of S 0 or S 00 , so we are done by (5.15) and (5.16).
Proof of bound in step j ∈ [2, N − 1]; j = j 0 6= j 00 . Since j 0 > 0, we have
= 1. There are three cases to consider: (i) S 0 | %1,j , S 0 - πj ; (ii) S 0 | πj ,
S 0 - %1,j , S 0 | %i,j with ji < j; (iii) S 0 | πj , S 0 - %1,j , S 0 | %i,j with ji = j; for
all cases suppose S 0 ∈ (U, 2U ] where U is a power of 2 in Ij . In case (i), (5.6)
gives
D1 = AS 0 (AS 0 − 4Pj 0 /m),
i0
with A = R1,j 0 /S 0 and Pj 0 /m fixed. Therefore, by (4.5), (4.9) and Lemma 4.2
(with k = 1, c = 1, Q = Qj 0 ),
X
X
U U <S 0 62U
X log Qj 0 (log log z)4
H1
¿ (log log x)5 (log x)δ .
¿
2
S0
log U
U
In case (ii), (5.6) gives
D1 = R1,j 0 (R1,j 0 − S 0 B),
with B = 4Pj /(mS 0 ) and R1,j 0 fixed. As in case (i), by Lemma 4.2 (with k = 1,
c = 1, Q = Qj 0 ), we have
X X H1
¿ (log log x)5 (log x)δ .
0
S
0
U U <S
62U
THE NUMBER OF SOLUTIONS OF φ(X)=M
309
For case (iii), we also have
D1 = R1,j 0 (R1,j 0 − S 0 B),
with B = 4Pj /(mS 0 ) and R1,j 0 fixed. Here we must sum over S 0 for which
S 0 R + 1 is prime, where R = Ri,j /S 0 is fixed. By (4.5) and Lemma 4.2 (with
k = 2, c = 1), we obtain
X log Qj 0 (log log x)4
X X H1
¿
¿ (log log x)5 (log x)δ−1+jδ .
3
0
S
log
U
U U <S 0 62U
U
The desired bounds now follow from (5.15) and (5.16).
A symmetric argument using (5.7) proves the step j bound when j =
j 00 6= j 0 .
Proof of bound in step j; j = j 0 = j 00 ; S 0 6= S 00 . By (5.3), ij−1 = 2. Define
by S 0 | %i∗ ,j , S 00 | %i∗∗ ,j . Put S 0 ∈ (U 0 , 2U 0 ], S 00 ∈ (U 00 , 2U 00 ] where U 0
and U 00 are powers of 2 in [ 12 Ej , Ej−1 ). By the Cauchy-Schwarz inequality,

1/2 
1/2
X X H1 H 2
X
X
X
1 
6
H12  
H22  .
0
00
0
00
S
S
U
U
0
00 0 00
0
00
0 00
0 00
i∗ ,
i∗∗
U ,U S ,S
U ,U
S ,S
S ,S
For each double sum over S 0 , S 00 , we first fix S 00 and sum over S 0 , then sum
on S 00 , bounding each sum with Lemma 4.2, Lemma 2.2 or the prime number
theorem, as appropriate. For example, for the sum on S 0 in the first sum we
use Lemma 4.2 (with c = 2) if D1 depends on S 0 . If i∗ > 3 then S 0 R0 + 1 is also
prime, where R0 = Ri∗ ,j /S 0 is fixed, so we use the k = 2 case of the lemma.
Otherwise we use the k = 1 case. If D1 does not depend on S 0 , we use Lemma
2.2 if i∗ > 3 and the prime number theorem otherwise. For the sum on S 00 ,
we use Lemma 4.2 only if D1 depends on S 00 but not on S 0 . If i∗∗ > 3 and
i∗∗ 6= i∗ , then we use the k = 2 case of Lemma 4.2, since S 00 R00 + 1 is prime,
where R00 = Ri∗∗ ,j /S 00 is fixed. Otherwise use the k = 1 case. If D1 does not
depend on S 00 , or it depends on both S 0 and S 00 , we apply either Lemma 2.2,
if i∗∗ > 3 and i∗∗ 6= i∗ , or the prime number theorem, otherwise. In every case
we obtain the bound
X
U 0 U 00 log4 Qj
H12 ¿
(log log x)10 ,
0 )k1 +3 (log U 00 )k2 +3
(log
U
0 00
S ,S
where k1 = 1 if i∗∗ > 3 and i∗∗ 6= i∗ , k1 = 0 otherwise, and k2 = 1 if i∗
k2 = 0 otherwise. The same bound holds for the sum on H22 , so
X X H1 H2
¿ (log log x)14 (log x)4δ−(k1 +k2 )(1−jδ) .
0 S 00
S
0
00 0 00
U ,U S ,S
The desired bound again follows at once from (5.15) and (5.16).
> 3 and
310
KEVIN FORD
Proof of bound for step j ∈ [2, N − 1]; j = j 0 = j 00 , S 0 = S 00 . Since
(πj , θj ) = (%1,j , %2,j ) = 1, either S 0 | πj , S 0 | %2,j or S 0 | θj , S 0 | %1,j . In either
case, by (5.6) and (5.7), D1 and D2 both depend on S 0 (recall τ and the Wi
for i ∈ I are fixed). As before we put S 0 ∈ (U, 2U ] where U is a power of 2 in
[ 12 Ej−1 , Ej ). By the Cauchy-Schwarz inequality and Lemma 4.2 (with c = 2,
k = 1) we obtain
!1/2 Ã
!1/2
Ã
X
X X H1 H2 X 1 X
2
2
6 U
H1
H2
S0
0
0
0
U U <S
62U
U
S
S
¿ (log log x) (log x) .
14
4δ
Combined with (5.15) and (5.16) gives the desired bound. This concludes the
proof of the bounds for steps j = 2 through j = N − 1.
Proof of the bound for step N . Let
∗
∗
τ = {πN
, θN
; %∗1,N , · · · , %∗M,N , w∗ },
where the starred variables (except w∗ ) are the quotients of the corresponding
∗ θ ∗ . We
unstarred variables and S 0 , S 00 , as appropriate. As before let t = πN
N
proceed as in steps 2 through N −1, except now there are no variables Wi to sum
over. The sum over S 0 and S 00 is handled exactly as in steps 2 through N −1, but
the sum over τ of 1/t is handled a bit differently. Since 2 ∈ IN , the procedure
in (5.16) will not work. By (4.11), we have Ω(t) 6 2δ(1 + δ) log log x + Ω(m).
As in the other steps, given t, the number of dual factorizations of t into M + 1
factors and 2 factors is at most (2M + 2)Ω(t) . We thus have
X1
X
1
¿ (2M + 2)2δ(1+δ) log log x
t
t
+
τ
P (t)6EN −1
¿ (log x)
δ+2δ(1+δ) log(2M +2)
,
and the bound for step N follows.
Acknowledgements. The author thanks Sergei Konyagin for helpful discussions and the referee for helpful criticisms of the exposition.
University of Texas at Austin, Austin, TX
Current address: University of South Carolina, Columbia, SC
E-mail address: ford@math.sc.edu
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(Received August 25, 1997)
(Revised April 28, 1998)